Question

Prove that a subgroup of a cyclic group is cyclic.

Answer 1

Theorem Every subgroup of a cyclic group is cyclic as well. Proof: Suppose that G is a cyclic group and H is a subgroup of G. Let g be the generator of G, G = {gn : n ∈ Z}. Denote by k the smallest positive integer such that gk ∈ H (if there is no such integer then H = {e}, which is a cyclic group).

Answer 2

Proof:- Let (G, o) be a cyclic group.
Then there exists a∈G  such that
       G = [a ] = {aⁿ : n∈Z}

Let H be a sbugroup of the group (G,o) then we have to show hat H is cyclic group.

If H = G or H = {e}
  then there is nothing to prove as  H is obviously cyclic.

Now consider H ≠ {e} , then every element of H is integral power of a.
if $a^r$∈H then $a^{-r}$∈H ( because H is a subgroup)

Thus H contains +ve as well as -ve integral power of a.
then there exists l∈H  such that,
                    $a^l$∈H
Now consider a set
 S = {n∈N : aⁿ∈H} ⊆ N
S  is a non empty subset of N, So by well ordering principal of set N; S has least element say m.

So m is least +ve integer such that $a^m$∈H
⇒ [$a^m$] ⊆ H                ............i)

We claim that  [$a^m$] = H
For ,
let x∈H
 ⇒ x∈H
 ⇒ x = aⁿ for some n∈Z
by division algorithm, there exists q , r ∈ Z  such that
n = mq + r     where 0≤r<m
⇒ r = n + m(-q)
$\implies\;a^r=a^{n+m(-q)}\\\;\\\implies\;a^r=a^n\;o\;a^{m(-q)}\\\;\\\implies\;a^r=a^n\;o\;a^{m(-q)}\in H\;\;\;\text{(Since H is closed)}$
$\implies\;a^r\in H$
and  0≤r<m,
So r must be equal to zero
   r = 0
thus n = mq
$\implies\;a^n=a^{mq}=(a^{m})^q\in[a^m]$
$\implies\;x\in[a^m]$
$x\in H\implies x\in[a^m]$
H ⊆ [$a^m$]    .............ii)
From i) and ii)
H =  [$a^m$]
Thus  H  is a cyclic group.

Note:- 1) A group is called a cyclic group if it is generated by a single element of a group, say a. Then  if group is cyclic, we can write all the elements of G as a integral power of 'a'

2) Every cyclic group is abelian group.